Chapter 11.2, Problem 7E

To determine

To find: the limit and another function which agrees the given function.

Answer to Problem 7E

  $\underset{x\to 1}{\lim }g\left(x\right)=2$

  $\underset{x\to -1}{\lim }g\left(x\right)=0$

  $\underset{x\to 0}{\lim }g\left(x\right)=0$

  $\underset{x\to -3}{\lim }g\left(x\right)=6$

The another function is $x\left(x+1\right)$

Explanation of Solution

Given:

  $g\left(x\right)=\frac{x^3-x}{x-1}$

  

Calculation:

(a)From the graph, $g\left(x\right)\to 2$ as $x\to 1$ from both the positive and negative directions, even though $g\left(x\right)$ is not defined there. Therefore $\underset{x\to 1}{\lim }g\left(x\right)=2$

(b) From the graph, $g\left(x\right)\to 0$ as $x\to -1$ from both the positive and negative directions. Also, find this by direct substitution, since $g\left(x\right)$ is a rational function with a non-zero denominator for $x=-1$ , therefore $\underset{x\to -1}{\lim }g\left(x\right)=0$

(c) $g\left(x\right)\to 0$ as $x\to 0$ from both directions, and it appears as if it is also defined to be exactly 0 at $x=0$ . Verify this by direct substitution, since $g\left(x\right)$ is a rational function with a non-zero denominator for $x=0$ . Therefore $\underset{x\to 0}{\lim }g\left(x\right)=0$

(d) $g\left(x\right)\to 6$ as $x\to -3$ from both directions,

  $\begin{array}{l}\underset{x\to -3}{\lim }g\left(x\right)=\frac{x^3-x}{x-1}\\ =\frac{{(-3)}^3+3}{-3-1}\\ =6\end{array}$

In order to identify another function that agrees with $g\left(x\right)$ at all but one point, use the "Dividing Out" technique described at the beginning of the chapter. First factor the numerator, then divide out any common factors between the numerator and denominator.

  $\begin{array}{l}\frac{x^3-x}{x-1}=\frac{x\left(x^2-1\right)}{x-1}\\ \text{ = }\frac{x\left(x-1\right)\left(x+1\right)}{x-1}\\ \text{ = }x\left(x+1\right)\end{array}$

Also, notice that this technique can be used to find the limit from part (a), since $g\left(1\right)=0/0$ , which is indeterminate, but by plugging $x=1$ into this function that agrees with $g\left(x\right)$ at all but one point, the limit can be found.

Conclusion:

Therefore,

  $\underset{x\to 1}{\lim }g\left(x\right)=2$

  $\underset{x\to -1}{\lim }g\left(x\right)=0$

  $\underset{x\to 0}{\lim }g\left(x\right)=0$

  $\underset{x\to -3}{\lim }g\left(x\right)=6$

The another function is $x\left(x+1\right)$